JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Critical Thinking

  • question_answer
    Two identical balls A and B each of mass 0.1 kg are attached to two identical massless springs. The spring mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in the figure. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius 0.06 m. Each spring has a natural length of 0.06p m and force constant 0.1N/m. Initially both the balls are displaced by an angle \[\theta =\pi /6\] radian with respect to the diameter \[PQ\] of the circle and released from rest. The frequency of oscillation of the ball B is 

    A)            \[\pi \,Hz\]

    B)            \[\frac{1}{\pi }Hz\]            

    C)            \[2\pi \,Hz\]

    D)            \[\frac{1}{2\pi }Hz\]

    Correct Answer: B

    Solution :

                       As here two masses are connected by two springs, this problem is equivalent to the oscillation of a reduced mass \[{{m}_{r}}\] of a spring of effective spring constant.                          \[T=2\pi \sqrt{\frac{{{m}_{r}}}{{{K}_{eff.}}}}\]                    Here \[{{m}_{r}}=\frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\frac{m}{2}\] Þ \[{{K}_{eff.}}={{K}_{1}}+{{K}_{2}}=2K\]            \ \[n=\frac{1}{2\pi }\sqrt{\frac{{{K}_{eff.}}}{{{m}_{r}}}}=\frac{1}{2\pi }\sqrt{\frac{2K}{m}\times 2}\]           \[=\frac{1}{\pi }\sqrt{\frac{K}{m}}=\frac{1}{\pi }\sqrt{\frac{0.1}{0.1}}=\frac{1}{\pi }Hz\]


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